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Maybe I'm a little rusty but I couldn't find the segment $OE$

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At the first sight, I thought the triangles EDG and EHA are similar, but it seems they aren't. If I remember well the answer in the end of the book is $9.025$.

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    $\begingroup$ ABE <---> BAE is a symmetry. Under that symmetry, C <---> D, and angle EGC <---> EGD, making both angles be 90 degrees. That gives me 8,375 for OE, but I could well be wrong. On the other hand, I also get that EH is 9.025, so maybe there's a typo in the problem or solution book. $\endgroup$ – John Hughes Sep 4 '17 at 0:36
  • $\begingroup$ @JohnHughes ABE and BAE are the same triangle, no? $\endgroup$ – user42912 Sep 4 '17 at 0:38
  • $\begingroup$ @JohnHughes Did you mean ABE and DEC? $\endgroup$ – user42912 Sep 4 '17 at 0:40
  • $\begingroup$ They are the same triangle, but with the vertices listed in a different order. This trick is, in a different context, known as the pons asinorum, I believe. $\endgroup$ – John Hughes Sep 4 '17 at 0:59
  • $\begingroup$ Nope. I meant exactly the things I said. I didn't give the whole argument by which I got my answer; I just corrected your misconception that EDG and EHA are not similar by showing that angle EDG is in fact 90 degrees. $\endgroup$ – John Hughes Sep 4 '17 at 1:00
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Note that the line through center perpendicular to chord bisects that chord (i.e. AH = HB). By SAS, we have $\triangle EAH \cong \triangle EBH$. Then, $\angle a1 = \angle a2$ and $\angle b = \angle b’$.

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Therefore, $\angle a4 = \angle a2$ ...(ext. angle cyclic quadrilateral)

$= \angle a1 = \angle a3$ ...(ext. angle cyclic quadrilateral)

This in turn means DC // AB, $\angle DGE = 90^0$, and therefore, $\triangle EDG \sim \triangle EAH$.

Then, $GH = \dfrac {GE \times AD}{DE} = …$, $OG = …$ and finally, $OE = …$

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  • $\begingroup$ Your argument could use a bit more detail. For instance: Because $\overleftrightarrow{EO}$ contains a diameter perpendicular to chord $\overline{AB}$, that diameter (and therefore, that line) bisects that chord. Therefore, by SAS, etc, etc. Also, it's a bit of a leap to conclude $\angle a4 = \angle a2 = \angle a1 = \angle a3$; of course, this follows readily from, say, the fact that opposite angles of a cyclic quadrilateral are supplementary (or something less sophisticated, if this result is unknown to OP), but you should be explicit. $\endgroup$ – Blue Sep 4 '17 at 5:46
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    $\begingroup$ @blue detail added in bold. $\endgroup$ – Mick Sep 4 '17 at 6:06

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